[EM] The Equal Vote Coalition and robla

[Richard]

On 5/19/25 12:21, Kristofer Munsterhjelm wrote:
 > My (implied) point is that the conference-hall method, and its
 > simulation, leaves something to be desired because additional voters
 > with an equal preference may be deprived of a say in the outcome. And if
 > it does leave something to be desired, it cannot unconditionally be
 > called "the correct" way to count.  ...

The counting method you describe (copied below) is also correct. 
Mathematically anyway.  It's more difficult for a typical voter to 
understand, especially because it involves fractions.  Yet it's correct. 
  And it yields the same result as the conference-hall method I explained.

The counting methods that are NOT correct are the only two options 
currently offered in the RCTab software at the Ranked Choice Voting 
Resource Center (RCVRC)!

One of those options is to skip over an "overvote", which is two or more 
candidates marked at the same preference level.  This is the option that 
was chosen for Portland (OR).  The other option is to dismiss ALL the 
remaining marks on the ballot when a so-called "overvote" is reached.

As I've said before, I'm hoping the RCVRC adds the third option of 
correctly counting so-called "overvotes."  Either your method (using 
fractions) or the conference-hall method (using pairings) would be the 
correct way to do the counting.

In the distant future, the rounding down of fractions, or dismissing 
unpaired ballots, could be omitted, but voters aren't yet willing to 
accept that.  Worse, critics of ranked choice voting would use the 
details to claim the counts are not completely correct.

As a reminder of why this issue matters, correct counting of "overvotes" 
will prevent the complications that arose in the recent Portland mayoral 
election using IRV.  Lots of us wanted to rank one or two frontrunners 
under lots of the other candidates, yet we were limited to marking only 
six ovals in six columns, with no more than one oval per column.  As a 
reminder there were about 20 candidates.

Also, when this counting issue is corrected, the promoters of STAR 
voting will have only one remaining valid criticism of IRV.  (And that 
weakness can be overcome by eliminating pairwise-losing candidates when 
they occur.)

The conference-hall explanation is intended for typical voters.  It 
helps them understand that an "overvote" is not a violation of "one 
person one vote."

As your example shows, if there is a change in elimination order among 
some less-popular candidates, that would not be likely to affect who wins.

In the meantime, governmental elections typically yield different vote 
counts each time the ballots are counted during an audit.  So counting 
correctly in a way that ignores twenty or so ballots during one of the 
counting rounds is a tiny unfairness compared to the current unfairness 
of the two methods supported by the RCTab software at the RCVRC, either 
of which can involve dismissing hundreds or thousands of ballots.

I continue to hope that Rob ("robla") can convey this insight to the 
folks at the Equal Vote Coalition so they recognize it's easy to 
overcome the two significant weaknesses of IRV.

To avoid any misunderstanding, I dislike IRV.  Yet its two significant 
weaknesses are easy to overcome without leaping to the complexity of 
Condorcet methods, and without abandoning the already-familiar idea of 
eliminating the least-popular candidate during each counting round.

Richard Fobes



On 5/19/25 12:21, Kristofer Munsterhjelm wrote:
> On 2025-04-24 18:41, Richard wrote:
>> On 4/24/25 02:20, Kristofer Munsterhjelm wrote:
>>
>>> What would happen if there were three such voters, all voting B=D? To be
>>> fair, it would seem you would have to place one and a half voter in B's
>>> line and the other in D's. But your constraints say you can't divide the
>>> voters into fractions.
>>
>> If the conference-hall simulation matched actual election counting 
>> rules, the third voter would not vote in that round.
>>
>> If a fourth voter also has the B=D equal preference, then that's two 
>> pairs of voters cooperating.
>>
>> If decimal numbers were used, the counts would be rounded down to the 
>> nearest integer.  This correctly simulates the pairing approach. 
>> Specifically, with three voters having the B=D preference, the count 
>> of three would be divided by the number of candidates involved, which 
>> is 2 (B and D), that yields 1.5 and that's rounded down to 1, which is 
>> the number of votes going to candidate B and the number of votes going 
>> to candidate D.
> 
> This seems problematic, though. In your original post, you said "Here's 
> how to correctly count overvotes in governmental elections, without 
> departing from the principle of one voter supporting only one candidate 
> in each round of voting."
> 
> My (implied) point is that the conference-hall method, and its 
> simulation, leaves something to be desired because additional voters 
> with an equal preference may be deprived of a say in the outcome. And if 
> it does leave something to be desired, it cannot unconditionally be 
> called "the correct" way to count. If "the principle of one voter 
> supporting only one candidate" is meant to be a constraint, it might be 
> (given the constraint), but then it remains to be shown that this 
> condition has value in itself.
> 
> As an example of uneven voting power, say there's a near tie election 
> with the following first preference counts so far:
> 
> A: 1700
> B: 1500
> C: 1300
> D: 1202
> E: 1200
> 
> and suppose that three ballots that rank A=E are counted next, then two 
> ballots ranking A=B=E, and finally three ballots ranking A=B=C=E.
> 
> By the conference-hall method, the three A=E ballots count for 1 point 
> for each:
> 
> A: 1701
> B: 1500
> C: 1300
> D: 1202
> E: 1201
> 
> and the two A=B=E ballots count for zero (rounding down), as do the 
> three A=B=C=E ballots. Hence, as far as I understand, E is eliminated.
> 
> By fully fractional counting, instead the three A=E ballots in total 
> count as 1 + 1/2 for each candidate. To mitigate roundoff problems, I'll 
> keep a table of how many of each fraction has been counted:
> 
>     wholes  /2   /3   /4   /5
> A: 1701    1    0    0    0
> B: 1500    0    0    0    0
> C: 1300    0    0    0    0
> D: 1202    0    0    0    0
> E: 1201    1    0    0    0
> 
> The two A=B=E ballots count, in total, as 2/3 for each:
> 
>     wholes  /2   /3   /4   /5
> A: 1701    1    2    0    0
> B: 1500    0    2    0    0
> C: 1300    0    0    0    0
> D: 1202    0    0    0    0
> E: 1201    1    2    0    0
> 
> and the three A=B=C=E ballots count as 3/4 for each:
> 
>     wholes  /2   /3   /4   /5
> A: 1701    1    2    3    0
> B: 1500    0    2    3    0
> C: 1300    0    0    3    0
> D: 1202    0    0    0    0
> E: 1201    1    2    3    0
> 
> Then D is eliminated instead of E because 1201 + 1/2 + 2/3 + 3/4 > 1202.
> 
>> The difference between what I'm suggesting and what you're suggesting 
>> is when division occurs.  I'm suggesting it happens once, after 
>> counting the ballots that share the same pattern.  You're suggesting 
>> that a ballot can be split into 0.5 votes for candidate B and 0.5 
>> votes for candidate D without waiting to see if there are an odd 
>> number of voters with that ranking pattern.  This means your approach 
>> does division at each ballot, instead of just once at the end of a 
>> counting round.
> 
> If by "division" you mean the splitting of ballots into fractions, 
> that's correct. But if you mean the mathematical operation of division, 
> then not necessarily. Consider the tabular form above, and let's look at 
> D and E here:
> 
>     wholes  /2   /3   /4   /5
> D: 1202    0    0    0    0
> E: 1201    1    2    3    0
> 
> Until this point, we haven't done any division, we've just been counting 
> fractions. Now we can do a division at the very end and find out that 
> 1201 + 1/2 + 2/3 + 3/4 > 1202.
> 
> This way of counting lends itself to easy reduction while the ballots 
> are counted, because once a candidate's count of halves (/2 fractions) 
> equals 2, then you can set it back to zero and increment the wholes/ 
> units place. Same with thirds (once you get three, just increment a unit 
> and set the thirds' count to zero).
> 
> This "increment the units" strategy has a similar effect as your 
> approach: two A=B voters contribute one point to each, in this case 
> because each adds a 1/2 fraction to both candidates, and then once you 
> get 2/2, you can increment the unit. In addition, you can keep a proper 
> count without having to keep track of how many e.g. B=C ballots you've 
> seen before when you encounter another one. Instead of thinking "I last 
> gave a point to C when I saw a B=C ballot, next one must be to B", you 
> can just add one /2 point to each, every time, and then resolve into 
> full votes once enough fractions accumulate.
> 
> The difference is that this tabular method keeps the residual fractions 
> until the end of the process so ties can be broken if necessary.
> 
> If the full vote values are too far apart, then you don't need to do a 
> division at all. E.g. if the result had been:
> 
>     wholes  /2   /3   /4   /5
> D: 1250    0    0    0    0
> E: 1201    1    2    3    0
> 
> then since every fractions column contributes less than a full vote 
> (otherwise, that full vote would have been drawn into the units place), 
> the value for E can't be greater than its unit count plus the number of 
> fractions columns, here 1201 + 4. Since 1205 < 1250, there's no need to 
> check the division: E would be the loser anyway.
> 
> This might be more complex, but it doesn't seem like the conference-hall 
> method is unambiguously "the correct way" to do things.
> 
> -km